Given two independent events, if the probability that exactly one of them occurs is `26/49` and the probability that none of them occurs is `15/49`, then the probability of more probable of the two events is :

Options

A. `4/7`
B. `6/7`
C. `3/7`
D. `5/7`

Answer

To solve the problem step by step, let's denote the probabilities of the two independent events A and B as P(A) and P(B).

### Step 1: Set Up the Given Information
We know:
- The probability that exactly one of the events occurs is \( P(A \text{ only}) + P(B \text{ only}) = P(A) \cdot (1 - P(B)) + (1 - P(A)) \cdot P(B) = \frac{26}{49} \).
- The probability that none of them occurs is \( P(\text{none}) = (1 - P(A)) \cdot (1 - P(B)) = \frac{15}{49} \).

### Step 2: Express the Probabilities
From the second piece of information, we can express:
\[ (1 - P(A))(1 - P(B)) = \frac{15}{49} \]
Let \( P(A) = x \) and \( P(B) = y \). Then:
\[ (1 - x)(1 - y) = \frac{15}{49} \]

### Step 3: Use the First Equation
From the first equation:
\[ P(A) \cdot (1 - P(B)) + (1 - P(A)) \cdot P(B) = \frac{26}{49} \]
This can be rewritten as:
\[ x(1 - y) + (1 - x)y = \frac{26}{49} \]
Expanding this gives:
\[ x - xy + y - xy = \frac{26}{49} \]
Which simplifies to:
\[ x + y - 2xy = \frac{26}{49} \]

### Step 4: Substitute and Solve
Now we have two equations:
1. \( (1 - x)(1 - y) = \frac{15}{49} \)
2. \( x + y - 2xy = \frac{26}{49} \)

From the first equation, we can expand:
\[ 1 - x - y + xy = \frac{15}{49} \]
Rearranging gives:
\[ xy - x - y = \frac{15}{49} - 1 = \frac{15 - 49}{49} = -\frac{34}{49} \]
So:
\[ xy - x - y = -\frac{34}{49} \]

### Step 5: Set Up a System of Equations
Now we have:
1. \( xy - x - y = -\frac{34}{49} \)
2. \( x + y - 2xy = \frac{26}{49} \)

From the first equation, we can express \( xy \) as:
\[ xy = x + y - \frac{34}{49} \]

### Step 6: Substitute into the Second Equation
Substituting \( xy \) into the second equation:
\[ x + y - 2(x + y - \frac{34}{49}) = \frac{26}{49} \]
This simplifies to:
\[ x + y - 2x - 2y + \frac{68}{49} = \frac{26}{49} \]
\[ -x - y + \frac{68}{49} = \frac{26}{49} \]
Rearranging gives:
\[ -x - y = \frac{26}{49} - \frac{68}{49} = -\frac{42}{49} \]
Thus:
\[ x + y = \frac{42}{49} = \frac{6}{7} \]

### Step 7: Find \( P(A) \) and \( P(B) \)
Now we have:
1. \( x + y = \frac{6}{7} \)
2. \( xy = \frac{8}{49} \) (from \( P(A \cap B) = P(A)P(B) \))

Let \( x \) and \( y \) be the roots of the quadratic equation:
\[ t^2 - (x+y)t + xy = 0 \]
Substituting the values:
\[ t^2 - \frac{6}{7}t + \frac{8}{49} = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{\frac{6}{7} \pm \sqrt{\left(\frac{6}{7}\right)^2 - 4 \cdot 1 \cdot \frac{8}{49}}}{2} \]
Calculating the discriminant:
\[ \left(\frac{6}{7}\right)^2 - 4 \cdot \frac{8}{49} = \frac{36}{49} - \frac{32}{49} = \frac{4}{49} \]
Thus:
\[ t = \frac{\frac{6}{7} \pm \frac{2}{7}}{2} = \frac{8/7}{2} \text{ or } \frac{4/7}{2} = \frac{4}{7} \text{ or } \frac{2}{7} \]

### Step 9: Identify the More Probable Event
The probabilities are \( P(A) = \frac{4}{7} \) and \( P(B) = \frac{2}{7} \). The more probable event is:
\[ \text{Probability of more probable event} = \frac{4}{7} \]

### Final Answer
The probability of the more probable of the two events is \( \frac{4}{7} \).

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